Generator-offset property: Difference between revisions

Line 34:

# ''S'' is ''pairwise-mos'' (PMOS). That is, the result of identifying any two step sizes of ''S'' is always a mos.

# ''S'' is ''monotone-mos'' (MMOS). That is, each of the following operations results in a mos: setting L = M, setting M = s, and setting s = 0.

# ''S'' is a billiard scale.

In particular, odd GO scales always satisfy these properties (see Proposition 2 below).

Line 82:

Line 83:

The generator of aX 2bW must have an odd number of W steps; if it had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction. This with (4) immediately gives (5).

For (6) and (7), we know SGA scales are rank-3 billiard scales and that rank-3 billiard scales project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW by (5). It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z. Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. Then g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply use the inverted generator class. Assume that the offset g1 = δ has one fewer Y (and one more Z) than g2. Then we use the inverted generator class iff g3 has one ''more'' X than both g1 and g2. <math>\square</math>

For (6) and (7), we know SGA scales are rank-3 billiard scales (why?) and that rank-3 billiard scales project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW by (5). It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z. Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. Then g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply use the inverted generator class. Assume that the offset g1 = δ has one fewer Y (and one more Z) than g2. Then we use the inverted generator class iff g3 has one ''more'' X than both g1 and g2. <math>\square</math>

=== Proposition 2 (Odd GO scales are SGA) ===

@@ Line 34: / Line 34: @@
 # ''S'' is ''pairwise-mos'' (PMOS). That is, the result of identifying any two step sizes of ''S'' is always a mos.
 # ''S'' is ''monotone-mos'' (MMOS). That is, each of the following operations results in a mos: setting L = M, setting M = s, and setting s = 0.
+# ''S'' is a billiard scale.
 In particular, odd GO scales always satisfy these properties (see Proposition 2 below).
@@ Line 82: / Line 83: @@
 The generator of aX 2bW must have an odd number of W steps; if it had an even number of W steps, it would be generated by stacking the generator of the mos aX bW' with W' = 2W, a contradiction. This with (4) immediately gives (5).
-For (6) and (7), we know SGA scales are rank-3 billiard scales and that rank-3 billiard scales project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW by (5). It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z. Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. Then g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply use the inverted generator class. Assume that the offset g1 = δ has one fewer Y (and one more Z) than g2. Then we use the inverted generator class iff g3 has one ''more'' X than both g1 and g2. <math>\square</math>
+For (6) and (7), we know SGA scales are rank-3 billiard scales (why?) and that rank-3 billiard scales project to mosses when one removes all instances of one step size, in particular s. Thus it suffices to prove that the result of equating any two step sizes is a mos. Equating Y and Z equates the swung alternants, resulting in the mos aX 2bW by (5). It remains to prove that equating X and Y equates the generator with the difference between the start of the 2nd chain with the end of the 1st one or vice versa; similarly for equating X and Z. Consider the swung-generator-alternant chain g1 g2 .... g1 g2 g3. Assume that the offset g1 = δ has one more Y (and one fewer Z) than g2 = g - δ. Then g3 (which becomes the imperfect generator of aX 2bW) either has one more X or one fewer X than both g1 and g2. If g3 has one more X, then g3 + g1 is equated to g1 + g2 = g, as desired. If g3 has one fewer X, then simply use the inverted generator class. Assume that the offset g1 = δ has one fewer Y (and one more Z) than g2. Then we use the inverted generator class iff g3 has one ''more'' X than both g1 and g2. <math>\square</math>
 === Proposition 2 (Odd GO scales are SGA) ===